Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Weinberg, QTF, pg 60:very specific question

  1. Jan 29, 2010 #1
    I have been able to read through up to here (middle of pg. 60). The instructions given are not clear: (I arranged the text so as to be easier to read on the page here.)

    "Next, let's apply rules (2.4.8), (2.4.9)

    Eq. (2.4.8): [tex]U(\Lambda,a)J^{\rho\sigma}U^{-1}(\Lambda,a) = \Lambda_{\mu}^{\rho}\Lambda_{\nu}^{\sigma}(J^{\mu\nu} - a^{\mu}P^{\nu} + a^{\nu}P^{\mu})[/tex]

    Eq. (2.4.9): [tex]U(\Lambda,a)P^{\rho}U^{-1}(\Lambda,a) = \Lambda_{\mu}^{\rho}P^{\mu}[/tex]

    to a transformation that is itself infinitesimal, i.e.,

    [tex]\Lambda^{\mu}_{\nu} = \delta^{\mu}_{\nu} + \omega^{\mu}_{\nu}[/tex]


    [tex]a^{\mu} = \epsilon^{\mu}[/tex]...

    Using Eq. (2.4.3),

    Eq. (2.4.3): [tex]U(1 + \omega , \epsilon) = 1 + \frac{1}{2} i \omega_{\rho\sigma} J^{\rho\sigma} - \epsilon_{\rho} P^{\rho} + . . . [/tex]

    and keeping only terms of first order in [tex]\omega[/tex] and [tex]\epsilon[/tex] Eqs. (2.4.8) and (2.4.9) now become

    Eq. (2.4.10): [tex]i [\frac{1}{2} \omega_{\mu\nu} J^{\mu\nu} - \epsilon_{\mu} P^{\mu} , J^{\rho\sigma}] = \omega_{\mu}^{\rho} J^{\mu\sigma} + \omega_{\nu}^{\sigma} J^{\sigma\nu} - \epsilon^{\rho} P^{\sigma} + \epsilon^{\sigma} P^{\rho}[/tex]


    Eq. (2.4.11): [tex]i [\frac{1}{2} \omega_{\mu\nu} J^{\mu\nu} - \epsilon_{\mu} P^{\mu} , P^{\rho}] = \omega_{\mu}^{\rho} P^{\mu}[/tex]."

    This may be a shot in the dark, but if I don't ask, I'll never know.

    Thanks in advance!
  2. jcsd
  3. Jan 29, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The instructions tell you to replace every [itex]a^\mu[/itex] in (2.4.8) and (2.4.9) with [itex]\epsilon^\mu[/itex], and every [itex]\Lambda^\mu{}_\nu[/itex] with [itex]\delta^\mu_\nu+\omega^\mu{}_\nu[/itex].

    This turns (2.4.8) into

    [tex](1+\frac i 2\omega_{\alpha\beta}J^{\alpha\beta}-i\epsilon_\alpha P^\alpha)J^{\rho\sigma}(1-\frac i 2\omega_{\gamma\delta}J^{\gamma\delta}+i\epsilon_\gamma P^\gamma)=(\delta^\rho_\mu+\omega^\rho{}_\mu)(\delta^\sigma_\nu+\omega^\sigma{}_\nu)(J^{\mu\nu}-\epsilon^\mu P^\nu+\epsilon^\nu P^\mu)[/tex]

    Looks like you forgot an i. I may have introduced some typos on my own though. These things have to agree order-by-order in each variable, just like two polynomials that are equal for all values of all the variables. (Wouldn't that mean that they're the same polynomial? Yes it would, that's the point). The zeroth order terms are

    [tex]J^{\rho\sigma}=\delta^\rho_\mu\delta^\sigma_\nu J^{\mu\nu}[/tex]

    Next, consider the terms that are of first order in components of [itex]\omega[/itex].

    (I have to leave the computer for a while, but I might post those details in a couple of hours).

    Also, make sure you understand the stuff in this post (about the notation) perfectly.


    [tex]\frac i 2\omega_{\alpha\beta}J^{\alpha\beta}J^{\rho\sigma}-J^{\rho\sigma}\frac i 2\omega_{\gamma\delta}J^{\gamma\delta}=\omega^\rho{}_\mu\delta^\sigma_\nu J^{\mu\nu}+\delta^\rho_\mu\omega^\sigma{}_\nu J^{\mu\nu}[/tex]

    [tex]\frac i 2\omega_{\alpha\beta}[J^{\alpha\beta},J^{\rho\sigma}]=\omega^\rho{}_\mu J^{\mu\sigma}+\omega^\sigma{}_\nu J^{\rho\nu}[/tex]

    Looks like you got an index wrong in the last term on the right.

    Weinberg kept all of the first order terms, so to get his result you repeat the above for terms of first order in [itex]\epsilon[/itex] and add the result to the result I got. Then do the same for (2.4.9).
    Last edited: Jan 29, 2010
  4. Jan 29, 2010 #3
    D'OH! I didn't associate the Unitary operators of (2.4.8) and (2.4.9), [tex]U(\Lambda,a)[/tex] and [tex]U^{-1}(\Lambda,a)[/tex], with (2.4.3)!

    I kept thinking that the [tex]\Lambda[/tex]'s were being replaced ("...itself...") That's what I was missing... the forest for the trees.

    And yes, I dropped an 'i' for the [itex]\epsilon_\rho P^\rho[/itex] term of (2.4.3).

    No need. I was confused by that one term only and the rest should be straight forward. I actually find the indexing to be helpful.

    That was another question I had! How could the 2 LT's go from a forward and an inverse boost to 2 inverse boosts?

    [tex]\emph{A million thanks!}[/tex]
    Last edited by a moderator: Apr 24, 2017
  5. Jan 31, 2010 #4
    [tex]U(\Lambda, a) \rightarrow U(1 + \omega, a) \approx 1 + \frac{i}{2}\omega_{\rho\sigma}J^{\rho\sigma} - i\epsilon_\rho P^\rho + ...[/tex]

    then how does one write

    [tex]U^{-1}(\Lambda, a)??[/tex]

    [tex]U^{-1}(\Lambda, a) \rightarrow U^{-1}(1 + \omega, a) \approx 1 - \frac{i}{2}\omega_{\rho\sigma}J^{\rho\sigma} + i\epsilon_\rho P^\rho + ... ???[/tex]


    [tex]{\Lambda^\mu}_\nu = \delta^\mu_\nu + {\omega^\mu}_\nu[/tex]


    [tex]{(\Lambda^{-1})^\mu}_\nu = {\Lambda_\nu}^\mu = \delta^\mu_\nu + {\omega_\nu}^\mu[/tex]

    But how is this incorporated into

    [tex]U^{-1}(\Lambda, a) ??[/tex]

    (Sorry, I thought I could do the details by myself.)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook